a, $4x^2+4xy+4y^2-4x+10y+13=0$

$(4x^2+y^2+1-4x+4xy-2y)+(3y^2-12y+12)=0$

$(2x+y-1)^2+3(y^2-4y+4)=0$

$(2x+y-1)^2+3(y-2)^2=0$

$\left \{ {{(2x+y-1)^2=0} \atop {3(y-2)^2=0}} \right.$ ⇒$\left \{ {{2x+y-1=0} \atop {y-2=0}} \right.$  ⇒$\left \{ {{\frac{-1}{2} } \atop {y=2}} \right.$ 

b, $2y^2 + 2z^2 + x^2 − 2xy − 2yz + 2xz − 2z − 4y + 5 = 0$

$(x^2+y^2+z^2-2xy-2zy+2xz)+(z^2-2z+1)+(y^2-4y+1)=0$

$(x-y+z)^2+(z-1)^2+(y-2)^2=0$

Ta có : $(x-y+z)^2=0$ ⇒$x=1$

           $(z-1)^2=0$ ⇒$z=1$

           $(y-2)^2=0$ ⇒$y=2$